Finding equilibrium ratio A:B when K=2.3 and products are 70%

[HAF]

Summary:: Direction of reaction A = 2B and we know that there is 70% of products and K = 2.3

Question is: What is the direction of reaction A = 2B and we know that there is 70% of products. K = 2.3

K = B^2 / A

B:A = sqrt (K.A) / A

I have came here but what should I do now? Because when I change the concentration of A, the ratio changes too and thus I can't figure out how to calculate the ratio.

Thank you for help

 

[BvU]

when I change the concentration of A

Why would you want to do that ?

In your terrible notation, what is B^2/A ?

 

[mjc123]

First, if you are going to use square brackets for concentration, remember to use [B ] (with a space) for the concentration of B. Without the space, it is the command for bold type.

Are you familiar with the concept of the reaction quotient Q? It is like K, except you use the actual concentrations present in the mixture, so
Q = [B_act]²/[A_act]
whereas K = [B_eq]²/[A_eq]
At equilibrium Q = K. If Q < K, the reaction has to go to the right, to increase [B ] and decrease [A], until Q = K. Vice versa if Q > K.

But here's the thing. You can't calculate Q with the data available. If "70% products" means that there are 0.3n moles of A and 0.7n moles of B in a volume V, then
Q = (0.7n/V)²/(0.3n/V) = 1.633n/V, so Q varies with n/V. (This is the case whenever there are not the same number of moles on each side of the equation.) Unless the original question supplies you with more information, you can't solve it.